# What is the Lie algebra of the indefinite orthogonal group''?

For $p,q \geq 0$ and $n=p+q\geq 1$, give $\mathbb{R}^n$ the indefinite inner product (written as a matrix) $$\begin{pmatrix} I_p & \\ & -I_q \end{pmatrix},$$ where $I_m$ is the $m \times m$ identity matrix. For example, if $\{e_i\}$ is a basis of $\mathbb{R}^n$ and $X = X^i e_i,$ then $$|X|^2 = (X^1)^2 + \cdots + (X^p)^2 - (X^{p+1})^2 - \cdots - (X^{p+q})^2.$$

Let $\mathrm{O}(p,q,\mathbb{R})$ be the Lie group of all linear transformations $T : \mathbb{R}^n \rightarrow \mathbb{R}^n$ that preserve this indefinite inner product.

What is the Lie algebra of $\mathrm{O}(p,q,\mathbb{R})$? Does it admit a nice'' description when $p$ and $q$ are both positive?

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## 1 Answer

Write $$I_{p,q} = \begin{pmatrix} I_p & 0 \\ 0 & -I_q \end{pmatrix}.$$ Then the Lie algebra of $\mathrm{O}(p,q; \mathbb{R})$ is given by $$\mathfrak{so}(p,q) = \{X \in M_n(\mathbb{R}) : X^T I_{p,q} = -I_{p,q} X\}.$$

Note that if you similarly define the indefinite unitary group $\mathrm{U}(p,q)$, then its Lie algebra is $$\mathfrak{u}(p,q) = \{X \in M_n(\mathbb{C}) : X^\dagger I_{p,q} = -I_{p,q} X\}.$$

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Just to add--these are all non-compact real forms of the corresponding complex groups $O(n, \mathbb{C})$, etc. So from the point of view of their complex representations, the signature $(p,q)$ does not really play a role. But the signature does play a role in the real representation theory (see e.g. the classification of spinors, which exhibits mod 8 periodicity). –  Jonathan May 1 '12 at 0:03
I read somewhere that the dimension of the lie algebra is $n(n-1)/2$, where $p+q=n$, but I am not being able to prove it. It seems like a totally trivial thing, but I have no idea how you get it. –  user23238 Dec 11 '12 at 10:42
@ramanujan_dirac: the Lie algebra consists of $n$ by $n$ matrices $X$ solving $XI_{p,q}=-I_{p,q}X$. Equivalently: $X_{ij}=-X_{ji}$ for $1\leq i,j\leq p$ and $p<i,j\leq n$ while $X_{ij}=X_{ji}$ for $1\leq i\leq p<j\leq n$. So the diagonal entries must vanish, and of the remaining $n^2-n$, the lower triangular entries are determined by the $n(n-1)/2$ upper triangular ones... –  Fran Burstall Jul 16 '13 at 19:44